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barcode generator for ssrs If this constraint is satisfied, the output of the downsampler has a DTFT as shown below. in Software
If this constraint is satisfied, the output of the downsampler has a DTFT as shown below. Scanning Code 128 Code Set A In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. ANSI/AIM Code 128 Creator In None Using Barcode creation for Software Control to generate, create Code 128B image in Software applications. Going through the D/C converter produces the signal y o ( [ ) ,which has the Fourier transform shown below. Code 128 Decoder In None Using Barcode scanner for Software Control to read, scan read, scan image in Software applications. Code 128 Code Set B Creation In C# Using Barcode drawer for .NET framework Control to generate, create Code 128A image in VS .NET applications. Therefore, in order for y o ( / ) to be equal to x u ( / ) ,we require that
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Print Bar Code In ObjectiveC Using Barcode generator for iPhone Control to generate, create barcode image in iPhone applications. GS1 RSS Printer In Java Using Barcode generator for Java Control to generate, create GS1 RSS image in Java applications. Find the smallest possible values for L and M and find the appropriate filter H(eJu)to perform this conversion. Given that 48000 = 2' . 3 . 5' and 44100 = 22 . 32 . S2 . 72, to change the sampling rate we require
Therefore, if we upsample by L = 160 and then downsample by M = 147, we achieve the desired sample rate conversion. The lowpass filter that we require is one that has a cutoff frequency wc = min(Tr
= 160 and the gain of the filter should be equal to L = 160.
Suppose that we would like to slow a segment of speech to onehalf its normal speed. The speech signal s a ( t ) is assumed to have no energy outside of 5 kHz, and is sampled at a rate of 10 kHz, yielding the sequence s ( n ) = s,(nT,) The following system is proposed to create the sloweddown speech signal.
Assume that S a ( j Q ) is as shown in the following figure: (a) Find the spectrum of v ( n ) . (b) Suppose that the discretetime filter is described by the difference equation Find the frequency response of the filter and describe its effecl on v ( n ) . (c) What is Y a ( j Q ) in terms of X , ( j Q ) Does y o ( ! ) correspond to sloweddown speech
(a) Since so(,) is sampled at the Nyquist rate, the DTFT of the sampled speech signal, s(n), is as follows: WeJw) Upsampling by a factor of 2 scales the frequency axis of S(el") by a factor of two as shown below.
SAMPLING
[CHAP. 3
The unit sample response of the discretetime filter is
which has a frequency response
H ( e J W= I ) + cos w
To see the effect of this filter on v ( n ) ,note that due to the upsampling, v ( n ) = 0 for n odd. Therefore. with it follows that
y(n) = u(n iv(nI)+;u(n+l) n odd
neven
Thus, the evenindex values of v ( n ) are unchanged, and the oddindex values are the average of the two neighboring values. As a result, h ( n ) performs a linear interpolation between the values of v(n). The output of the DC converter, y,,(t), has a Fourier transform Y,(.W) = T T Y ( e J R T r ) IS21 < TIT, otherwise
Since and
Y (ej") = H (pi''')v (ei") = ( I
+ cos w
)(ej") ~
v ( e J w ) s(eJ2") = then which is the product of (1 + cos QT,) and T , S ( ~ J ' " ~as%illustrated below. ) Thus, y,(t) does not correspond to sloweddown speech due to the images of s,,(r) that occur in the frequency range 5000n < IS21 < IOOOOn and the nonideal linear interpolator. Note that a better approximation would be to use a DC converter with a sampling rate of 2T, to eliminate the images. Shown in the figure below are two different ways of cascading an upsampler with a downsampler.
CHAP. 31
SAMPLING
(a) If M = L , show that the two systems are not identical. (b) Under what conditions will the two systems be identical (a) In the first system, which consists of an upsampler followed by a downsampler, note that w l ( n )is a sequence that is formed by inserting L  I zeros between each value of x ( n ) . The downsampler then extracts every Lth value of w l ( n ) ,thus producing the output In the second system, however, the downsampler extracts every Lth sample of s ( n ) and discards the rest. The upsampler then inserts L  I zeros between each value of w 2 ( n j . Thus, Therefore, the two systems are not the same.
(b) In order to analyze these systems when L following figure: # M, note that y z ( n ) in the second system has the form shown in the
On the olher hand, the sequence w , ( n ) in the first system is as shown below.
Note that y l ( n ) is formed by extracting every Mth value of w l ( n ) , SAMPLING Clearly, so yl(kL) = wl(kML) = x(kM) yl(kL) = yz(kL)

